Question

A manometer using oil (density 0.900 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 7.28 mmHg. Density of mercury is 13.6 g/cm3. By how much does the fluid level rise in the side of the manometer that is open to the atmosphere?What would your answer be if the manometer used mercury instead?

Answer 1

Rise in level of fluid is 0.11 m

Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm

Solution:

As per the question:

Density of oil,
`\rho_(o) = 0.900\ g/cm^(3) = 900\ kg/m^(3)`

Change in Pressure in the tank,
`\Delta P = 7.28\ mmHg`

Density of the mercury,
`\rho_(m) = 13.6\ g/cm^(3) = 13600\ kg/m^(3)`

Now,

To calculate the rise in the level of fluid inside the manometer:

We know that:

1 mmHg = 133.332 Pa

Thus

`\Delta P = 7.28\ times 133.332 = 970.656\ Pa`

Also,

`\Delta P = \rho_(o) gh`

where

g = acceleration due to gravity

h = height of the fluid level

`970.656 = 900* 9.8* h`

h = 0.11 m

Now, if mercury is used:

`\Delta P = \rho_(m) gh`

`970.656 = 13600* 9.8* h`

h = 0.00728 m = 7.28 mm